An infinite non-conducting sheet has uniform surface charge density σ. At a point P at perpendicular distance x from the sheet, the field magnitude is E₀ = σ/(2ε₀). A circular portion of radius r, centred at the foot O of the perpendicular from P, is removed, where r ≪ x.
The reduction ΔE in the field at P is:

Question

An infinite non-conducting sheet has uniform surface charge density σ. At a point P at perpendicular distance x from the sheet, the field magnitude is E₀ = σ/(2ε₀). A circular portion of radius r, centred at the foot O of the perpendicular from P, is removed, where r ≪ x.

The reduction ΔE in the field at P is:

G1 Physics · Accepted Answer

The removed portion is a uniformly charged disc of radius r. Its axial field at distance x is E d i s c E_disc E d ​ i sc = (σ/2ε₀) [1 - x/√(x²+r²)]. Removing the disc subtracts exactly this contribution. Therefore, ΔE = (σ/2ε₀) [1 - x/√(x²+r²)]. For r ≪ x, ΔE ≈ σr²/(4ε₀x²).